Question

Calculate the mass of $Hg$ which can be prepared by the reduction of mercury (II) ions in the presence of chloride ions by the passage of $5.0$ ampere current for $3.0$ hours.

Answer 1

Reduction reaction is

$H{g}^{2+}+2e\to H{g}_{\left(s\right)}$

(i) mole ratio$=\frac{molesofHgformed}{molesofelectrons}=\frac{1}{2}$

(ii) Moles of $Hg$ produced$=\frac{I\times t\left(C\right)}{96500Cmo{l}^{-1}}\times \frac{1}{2}$

$\therefore \frac{MassofHg}{Atomicmass}=\frac{5\times 10800\left(C\right)}{96500Cmo{l}^{-1}}\times \frac{1}{2}$

$\therefore \frac{mass}{200.59}=\frac{5\times 10800}{96500}\times \frac{1}{2}$

$\therefore mass=\frac{5\times 10800\times 200.59}{96500\times 2}$

$mass=56.1g$

Mass of $Hg$ that can be prepared is $56.1g$.